As it turns out, a field endowed with a pre-positive cone has an orderstructure. The field is called a formally real, orderable, or ordered field. Before defining what this “order” is, let’s do some preliminary work. Let P0subscriptP0P_{0} be a pre-positive cone of a field FFF. By Zorn’s Lemma, the set of pre-positive cones extending P0subscriptP0P_{0} has a maximal elementPPP. It can be shown that PPP has two additional properties:

5.

P∪(-P)=FPPFP\cup(-P)=F

6.

P∩(-P)=(0).PP0P\cap(-P)=(0).

Proof.

First, suppose there is a∈F-(P∪(-P))aFPPa\in F-(P\cup(-P)). Let P¯=P+P⁢anormal-¯PPPa\overline{P}=P+Pa. Then a∈P¯anormal-¯Pa\in\overline{P} and so PPP is strictlycontained in P¯normal-¯P\overline{P}. Clearly, sqr⁡(F)⊆P¯sqrFnormal-¯P\operatorname{sqr}(F)\subseteq\overline{P} and P¯normal-¯P\overline{P} is easily seen to be additively closed. Also, P¯normal-¯P\overline{P} is multiplicatively closed as the equation(p1+q1⁢a)⁢(p2+q2⁢a)=(p1⁢p2+q1⁢q2⁢a2)+(p1⁢q2+q1⁢p2)⁢asubscriptp1subscriptq1asubscriptp2subscriptq2asubscriptp1subscriptp2subscriptq1subscriptq2superscripta2subscriptp1subscriptq2subscriptq1subscriptp2a(p_{1}+q_{1}a)(p_{2}+q_{2}a)=(p_{1}p_{2}+q_{1}q_{2}a^{2})+(p_{1}q_{2}+q_{1}p_{%
2})a demonstrates. Since PPP is a maximal and P¯normal-¯P\overline{P} properly containsPPP, P¯normal-¯P\overline{P} is not a pre-positive cone, which means-1∈P¯1normal-¯P-1\in\overline{P}. Write -1=p+q⁢a1pqa-1=p+qa. Then q⁢(-a)=p+1∈Pqap1Pq(-a)=p+1\in P. Since q∈PqPq\in P, 1/q=q⁢(1/q)2∈P1qqsuperscript1q2P1/q=q(1/q)^{2}\in P, -a=(1/q)⁢(p+1)∈Pa1qp1P-a=(1/q)(p+1)\in P, contradicting the assumption that a∉-PaPa\notin-P. Therefore, P∪(-P)=FPPFP\cup(-P)=F.

For the second part, suppose a∈P∩(-P)aPPa\in P\cap(-P). Since a∈-PaPa\in-P, -a∈PaP-a\in P. If a≠0a0a\neq 0, then -1=a⁢(-a)⁢(1/a)2∈P1aasuperscript1a2P-1=a(-a)(1/a)^{2}\in P, a contradiction.
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A subset PPP of a field FFF satisfying conditions 1, 2, 5 and 6 is called a positive cone of FFF.
A positive cone is a pre-positive cone. If a∈FaFa\in F, then either a∈PaPa\in P or -a∈PaP-a\in P. In either case, a2∈Psuperscripta2Pa^{2}\in P.
Next, if -1∈P1P-1\in P, then 1∈-P1P1\in-P. But 1=12∈P1superscript12P1=1^{2}\in P, we have 1∈P∩(-P)1PP1\in P\cap(-P), contradicting Condition 6 of PPP.